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If \sin A = \frac{\sqrt{5}-1}{4} , then the value of  \frac{ \sin^2 A}{1-\cos A} is

(A lies in first quadrant)

Option: 1

1-\frac{\sqrt{5}-1}{4}


Option: 2

1+\frac{\sqrt{5}-1}{4}


Option: 3

=1-\frac{\sqrt{10+2\sqrt{5}}}{4}


Option: 4

1+\frac{\sqrt{10+2\sqrt{5}}}{4}


Answers (1)

best_answer

Clearly A = 18o

Now,

\\\frac{\sin^2 A}{1- \cos A}\\\\=\frac{1-\cos^2 A}{1- \cos A}\\\\ =\frac{(1-\cos A)(1+ \cos A)}{1- \cos A}\\ \\=1+\cos A\\\\ =1+\frac{\sqrt{10+2\sqrt{5}}}{4}

Posted by

Divya Prakash Singh

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